Triangle with Reverse Starting Letter in C

What You’ll Learn

Print an alphabet triangle where each row starts one letter earlier, but letters still run forward to a fixed top — E, DE, CDE, BCDE, ABCDE. Mixes ideas from Program 1 (forward run) and Program 2 (moving start). Includes a live preview, worked C examples, edge cases, and complexity.

📝 Problem & Approach

Given a row count n (or fixed A–E), print a left-aligned triangle of forward alphabet suffixes ending at a fixed top.

// Five rows (top = E)
// E
// DE
// CDE
// BCDE
// ABCDE

Inputs & Outputs

Minimal workflow

top = 'A' + rows - 1
for i from top down to 'A':      // start letter
    for j from i up to top:      // forward run
        print j
    print newline

Approach comparison

🔮 Live Preview

Choose 1–10 rows and draw the reverse-starting-letter alphabet triangle in the browser.

Press "Draw pattern".

Examples Gallery

Three complete C programs — fixed A–E, scanf row count, and a spaced-letter variant. Click View Output to reveal sample console results.

#include <stdio.h>

int main() {
    char i, j;

    for (i = 'E'; i >= 'A'; i--) {
        for (j = i; j <= 'E'; j++) {
            printf("%c", j);
        }
        printf("\n");
    }

    return 0;
}

E
DE
CDE
BCDE
ABCDE

#include <stdio.h>

int main() {
    int rows;
    char top, i, j;

    printf("Enter the number of rows: ");
    scanf("%d", &rows);

    top = (char)('A' + rows - 1);
    for (i = top; i >= 'A'; i--) {
        for (j = i; j <= top; j++) {
            printf("%c", j);
        }
        printf("\n");
    }

    return 0;
}

Enter the number of rows: 4
D
CD
BCD
ABCD

#include <stdio.h>

int main() {
    char top = 'E';
    char i, j;

    for (i = top; i >= 'A'; i--) {
        for (j = i; j <= top; j++) {
            printf("%c ", j);
        }
        printf("\n");
    }

    return 0;
}

E 
D E 
C D E 
B C D E 
A B C D E 

🔎 Worked Walkthrough — Top = E (5 rows)

Trace each start letter and the resulting forward suffix.

Row lengths are 1, 2, 3, 4, 5. The right edge is always E.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

For n rows you print 1+2+…+n = n(n+1)/2 letters, so total work is O(n²).

💡 Best Practices